Combinatorial Analysis
This section will cover counting, combinations, permutations, and urn models. 'Basic Counting Principle' If experiment 1 has m'' outcomes and experiment 2 has ''n outcomes, then the combined experiment has mn total outcomes. :: Example 1: Suppose you flip a fair coin and roll a fair 6-sided die. How many total possible combinations are there for (flip, roll) and what are the possible outcomes? :: Flip has 2 outcomes: {heads, tails} :: Roll has 6 outcomes: {1, 2, 3, 4, 5, 6} :: The total possible outcomes is 2 * 6 = 12 :: (H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6) 'Generalized Counting Principle' If experiment 1 has n1 outcomes and experiment 2 has n2 outcomes and so on and experiment k'' has nk outcomes, the combined experiment has n1 * n2 * … * nk'' total outcomes. :: Example 1: Suppose a committee is to be formed with 3 people: 1 student, 1 faculty member, and 1 staff. Suppose there are 9 students, 40 faculty, and 100 staff. How many different committees could be formed? :: 9 * 40 * 100 = 36,000 different possible committees :: Example 2: '''Suppose North Carolina lisence plates have 3 letters followed by 4 digits. How many different lisence plates are there? :: There are 26 letters in the alphabet and 3 possible locations for letters :: There are 10 digits and 4 possible locations for digits :: 26 * 26 * 26 * 10 * 10 * 10 * 10 = 175,760,000 possible combinations :: '''Example 3: What if you cannot repeat digits or letters in the license plate? :: 26 * 25 * 24 * 10 * 9 * 8 * 7 = 78,624,000 possible combinations without repeating digits or letters :: Example 4: Area codes used to have the property that the first digit couldn't be a zero or 1, the second had to be a zero or 1, and the third digit could be any digit. How many different area codes were possible? :: 8 * 2 * 10 = 160 possible area codes 'Permutations' The basic question of how many ways you can arrange things in a row. What if we have 2 people, A and B, how many combinations are there? AB or BA. If there is a third person, C, then they could be ABC, BAC, ACB, BCA, CAB, or CBA. You could have figured this out using the counting principle (3 * 2 * 1 = 6 different possible combinations). If we had n'' people, it's represented by ''n * (n''-1) * (''n-2) * ... * 2 * 1 = n''! So the number of ways to permute ''n objects in a row is n''! :: '''Example 1:' How many ways could 9 students sit in a row? :: 9! = 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 = 362,880 :: Example 2: Suppose Bill has 8 books on a shelf: 3 calculus, 2 algebra, 1 probability, and 2 linear algebra. :: a) How many ways could the books be arranged if there are no restrictions? :: 8! = 40,320 :: b) How many ways could the books be arranged if books from the sam area had to be grouped togther? :: This requires 2 steps: ::: i) What order will the 4 areas go in? 4! is the order of the areas ::: ii) What order will the books in a specific area appear? 3! * 2! * 1! * 2! is the order of the books within each area :: Then multiply them together to determine your answer: :: 4! * 3! * 2! * 1! * 2! = 576 :: Example 3: How many ways could you rearrange the letters in the word MATH? :: 4! = 24 :: Example 4: How many ways could you rearrange the letters in the word EGG? :: 3! / 2! = 6 / 2 = 3 In general, if you have n'' objects of which n1 are alike, n2 are alike, ..., nk are alike then there are: [''n! / (n1!'' *'' n2!'' * … * nk''!)] ways to permute them. :: Example 5: How many different arrangements are there for the letters in RACECAR? :: / (2! * 2! * 2! * 1!) = 630 :: Example 6: How many different ways are there to arrange the letters in STEELERS? :: / (2! * 1! * 3! * 1! * 1!) = 3360 :: Example 7: (a) How many ways could 9 students sit in a row if Andrew and Daniel have to sit next to each other? It's really like we only have 8 seats so there are 8! ways AND there are 2! ways to order Andrew and Daniel so: :: 8! * 2! = 80,640 :: (b) What if Andrew and Daniel refuse to sit next to each other? If there are no restrictions, there are 9! ways to do this. The number of ways they don't sit together is all possible arrangements minus Andrew and Daniel sitting next to each other, so: :: 9! - (8! * 2!) = 362,880 - 80,640 = 282,240 :: © What if Andrew and Daniel must have exactly 2 seats between them? Andrew and Daniel can sit 6 different spots and could switch so their position is given by 6 * 2! = 12 and everyone else can sit wherever so 7! then 12 * 7! = 60,480 Remember: With permutations, order matters! 'Combinations' We have a group of n'' objects and we want to choose ''k objects from the group. Order does not matter. :: Example 1: How many ways could we pick a group of 2 people from a group of 9 people? :: We have 9 choices for the first person and 8 choices for the second. Since order doesn't matter, we need to divide by 2! because every group of 2 people is actually picked twice with the permutation approach. The number of ways to choose 2 of the 9 people is: :: * 8) / (2 * 1) = 36 :: Also written as: :: * 8 * 7 * 6 * 5 * 4 * 3 * 2 *1) / (1 * 2 * 7 * 6 * 5 * 4 * 3 * 2 * 1) = / (2! * 7!) :: This give a formula with factorials in the notation 'Binomial Coefficient' The common notation of factorials is provided by the Binomial Coefficient. It represents the number of ways to choose k'' objects from a set of ''n objects. The equation is read as "n'' choose ''k equals..." Some important properties to keep in mind: : (a) 0! = 1 : (b) n'' choose 1 = ''n : © n'' choose k'' = zero if k'' > ''n or k'' < 0 :: '''Example 1:' We have a class of 9 people. :: (a) How many groups of 3 could we choose? :: (b) We have 5 women and 4 men in the class. What if we want a group of 3 with 2 women and 1 man? :: © What if we want 3 people but Mike and Kim will only be in a group together? Now we have to consider two parts: ::: i) What if Mike and Kim are in the group? :::: Choose both for the group (2 choose 2) :::: Choose one remaining person for the group (7 choose 1) :::: So that is (2 choose 2) * (7 choose 1) = 7 possibilities ::: ii) What if neither are in the group? :::: Choose zero if they are not in the group (2 choose 0) :::: Choose the 3 other people for the group (7 choose 3) :::: So that is (2 choose 0) * (7 choose 3) = 35 ::: iii) Add parts i and ii to find the answer: 7 + 35 = 42 different groups :: d) What if Mike and Kim refuse to be in a group together? :: This is the total wasy to a pick a group (9 choose 3) minus the groups with both Mike and Kim (7) so it's 77 possible groups :: Example 2: Powerball lottery. You have to pick 5 of the 55 white balls and one of the 42 red balls which is the power ball. How many different winning combinations are possible? :: So that's (55 choose 5) for the white balls times (42 choose 1) for the red balls :: (55 choose 5) * (42 choose 1) = 146,107,962 'Binomial Theorem' Let n'' be a natural number (''x+ y)n then the binomial theorem can be seen in the image to the left. Using the binomial theorem, you can calculate Pascal's Triangle. :: Example 1: Expand (x + y) ^ 3